1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Epsilon Delta Proof

  1. Jun 1, 2008 #1
    [SOLVED] Epsilon Delta Proof

    Does this limit proof make total sense? Given : "Show that [tex]\lim_{x \rightarrow 2} x^{2} = 4[/tex]."

    My attempt at it :[tex]0<|x^{2}-4|<\epsilon[/tex] which can also be written as [tex]0<|(x-2)(x+2)|<\epsilon[/tex].
    [tex]0<|x-2|<\delta[/tex] where [tex]\delta > 0[/tex]. It appears that [tex]\delta = \frac {\epsilon}{x+2}[/tex] which is the conversion factor. Which then by substitution, [tex]0<|x-2|<\frac {\epsilon}{x+2}[/tex].

    Here is the actual proof:
    Choose [tex]\delta = \frac {\epsilon}{x+2}[/tex], given [tex]epsilon > 0[/tex] then if [tex]0<|x-2|<\frac {\epsilon}{x+2}[/tex] then [tex]0<|(x-2)(x+2)|<\delta[/tex].

    Is this explanation coherent? I actually have a slight idea of what I wrote. Hopefully I am on the right path.
    Last edited: Jun 1, 2008
  2. jcsd
  3. Jun 1, 2008 #2


    User Avatar
    Homework Helper

    One of the more rigorous mathematicians here may have more to say about this then I will, but I think you have the gist of it. The one safeguard that is imposed in epsilon-delta proofs when you have a non-constant upper limit on the inequality is to make that limit

    min ( 1 , [tex]\frac {\epsilon}{x+2}[/tex] ) ,

    since the ratio can blow up at x = -2 and we've placed no restriction on the permitted values of x...
  4. Jun 1, 2008 #3
    I can't believe I forgot to mention that!
  5. Jun 1, 2008 #4
    When you are doing [tex]\epsilon-\delta [/tex] proofs, you have to say define what epsilon and delta are i.e. [tex]\forall \epsilon >0, \exists \delta[/tex] such that etc.
  6. Jun 1, 2008 #5
    Your [itex] \delta [/itex] should not depend on [itex] x [/itex]. As the post above mentions, the logic goes: for all [itex] \epsilon > 0 [/itex] there exists [itex] \delta > 0 [/itex] such that if [itex] |x-2| < \delta[/itex] then [itex] |x^2 - 4| < \epsilon [/itex]. Delta can, however, depend on epsilon. If |x-2| is bounded by [itex] \delta < 1 [/itex], then what is a bound on |x+2|?
  7. Jun 1, 2008 #6


    User Avatar
    Homework Helper

    Sorry -- quite so! It's been a while since I've looked at one of these proofs and I'm writing from someplace I don't have my books handy...
  8. Jun 2, 2008 #7
    My second attempt : Given any [tex]\epsilon > 0[/tex], choose [tex]\delta = \frac {\epsilon} {|x+2|} (x\neq -2)[/tex] where [tex]\delta > 0[/tex], then [tex]|x-2||x+2|<\epsilon[/tex] whenever [tex]|x-2|<\delta = \frac {\epsilon} {|x+2|}[/tex].

    Is this right?
    Last edited: Jun 2, 2008
  9. Jun 2, 2008 #8
    In my revision, I've been told that I did not prove anything. Is this a complete proof?
  10. Jun 2, 2008 #9
    Let [tex]f(x)=x^2[/tex], [tex]x_{0}=2[/tex]. [tex]\forall \epsilon>0,\exists \delta>0[/tex] such that [tex]|x^2-4|<\epsilon[/tex], [tex]|x-2|<\delta[/tex]
    Let [tex]\delta = \frac{\epsilon}{x+2}[/tex]
    ... continue from here. "..." [tex]<\epsilon[/tex]

    OR let [tex]\delta=min\{ 1,\frac{\epsilon}{5} \}[/tex] and use the triangle inequality.
    Last edited: Jun 2, 2008
  11. Jun 2, 2008 #10
    Substituting [tex]\delta = \frac {\epsilon}{x+2}[/tex] into [tex]|x-2|<\delta[/tex], we see that.. it works out? I do not know what to do from here.

    Nor do I know the min way.
  12. Jun 2, 2008 #11
    Continue from the proof,
    Let [tex]\epsilon>0[/tex], let [tex]\delta = min\{\frac{\epsilon}{5}\} [/tex]

    Assume that [tex]|x-2|<\delta[/tex]

    By Triangle inequality,
    [tex]|x^2-4|=|(x+2)(x-2)|\leq |x+2||x-2|[/tex]

    Since [tex]|x-2|<\delta[/tex], then [tex]|x-2|<\epsilon/5[/tex]

    Then for 0<x<3 such that [tex]|x+2|=5[/tex] **note: this is why you get the 5 as the denominator under epsilon.

    Thus, [tex]|x^2-4|\leq |x+2||x-2|<5*\frac{\epsilon}{5}=\epsilon[/tex] QED
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Epsilon Delta Proof
  1. Epsilon- Delta Proof (Replies: 2)